Question: $\dfrac{ 2q - 2r }{ 9 } = \dfrac{ 5q + 2s }{ -5 }$ Solve for $q$.
Multiply both sides by the left denominator. $\dfrac{ 2q - 2r }{ {9} } = \dfrac{ 5q + 2s }{ -5 }$ ${9} \cdot \dfrac{ 2q - 2r }{ {9} } = {9} \cdot \dfrac{ 5q + 2s }{ -5 }$ $2q - 2r = {9} \cdot \dfrac { 5q + 2s }{ -5 }$ Multiply both sides by the right denominator. $2q - 2r = 9 \cdot \dfrac{ 5q + 2s }{ -{5} }$ $-{5} \cdot \left( 2q - 2r \right) = -{5} \cdot 9 \cdot \dfrac{ 5q + 2s }{ -{5} }$ $-{5} \cdot \left( 2q - 2r \right) = 9 \cdot \left( 5q + 2s \right)$ Distribute both sides $-{5} \cdot \left( 2q - 2r \right) = {9} \cdot \left( 5q + 2s \right)$ $-{10}q + {10}r = {45}q + {18}s$ Combine $q$ terms on the left. $-{10q} + 10r = {45q} + 18s$ $-{55q} + 10r = 18s$ Move the $r$ term to the right. $-55q + {10r} = 18s$ $-55q = 18s - {10r}$ Isolate $q$ by dividing both sides by its coefficient. $-{55}q = 18s - 10r$ $q = \dfrac{ 18s - 10r }{ -{55} }$ Swap signs so the denominator isn't negative. $q = \dfrac{ -{18}s + {10}r }{ {55} }$